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Capacitance of 3 Hollow Cylinders

  1. Sep 27, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm trying to find the capacitance of a system of 3 concentric hollow cylinders. The first cylinder has radius R, the second radius 2R, the third radius 3R. Cylinders 1 and 3 are connected by a wire. In total, they have charge +λ, and the second cylinder has charge -λ. The cylinders are infinitely long.

    2. Relevant equations
    Ehollow cylinder = (1 / 2πεo)*(Q / RL)
    C= Q / V

    3. The attempt at a solution
    Since cylinders 1 and 3 are connected by the wire, the charge will distribute evenly according to surface area. Cylinder 3 has 3 times as much surface area, so it will have charge 3λ/4 (And Cylinder 1 will have charge λ/4).

    I know that to solve this problem I must integrate the electric field to find the potential, and use that in C=Q/V to find capacitance. However, I don't really understand what the limits of my integration are. The outer cylinders are positive, so their electric fields point inwards to the middle cylinder. Do I account for their opposite directions (cancelling each other out partially)? Do I just integrate E from R to 3R? Do I need to multiply two oppositely signed integrals by the proportion of charge?

    I was thinking of multiplying ∫E from R to 2R by λ/4, and multiplying ∫E from 2R to 3R by 3λ/4, then subtracting those. My believe my biggest trouble is that I don't understand how to deal with more than 2 surfaces for capacitor (and I can't find anything to help me online).
     
  2. jcsd
  3. Sep 27, 2015 #2

    Simon Bridge

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    Have you tried treating the setup as two capacitors?
     
  4. Sep 27, 2015 #3
    Thanks for the reply!
    Ok, so treating it as two capacitors (Where V1 is potential difference for the inner two cylinders and V3 is potential difference for the outer two cylinders):

    V1 = -∫R2R Q / 8πεLR dR = -Qln(2) / 8πεL = Qln(1/2) / 8πεL

    C= Q / V, so C1 = 8πεL / ln(1/2)

    Same thing for C3, except Q is multiplied by 3/4 instead of 1/4.
    C3 = 8πεL / 3ln(1/2)

    Do I add these together to get the cumulative capacitance of the system?
     
    Last edited: Sep 27, 2015
  5. Sep 28, 2015 #4
    Anyone know how I should proceed?
     
  6. Sep 28, 2015 #5

    Simon Bridge

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    Are the caps connected in series or parallel?
     
  7. Sep 28, 2015 #6
    I'm not sure I understand the question. The capacitor is 3 concentric cylinders, and the only wire in the problem is connecting the inner and outer cylinders (I guess this means they are connected in series?).
     
  8. Oct 5, 2015 #7

    Simon Bridge

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    You are modelling a single complicated capacitor as an electrically equivalent network of simpler ideal capacitors connected by ideal "wires".
    In this case you have two ideal caps - and there are two "wire" connections. (The ideal wires do not have to be physical wires - they can be any conductor that plays the role of equalizing the voltage between two places.)

    If you sketch out the series and parallel configurations for two ideal caps, and look at how the charges are distributed between the plates, you should be able to figure out the configuration that is right for you.
     
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